3b(b-2)=12

Solution for 3b(b-2)=12 equation:

3b(b-2)=12

We move all terms to the left:

3b(b-2)-(12)=0

We multiply parentheses

3b^2-6b-12=0

a = 3; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·3·(-12)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{5}}{2*3}=\frac{6-6\sqrt{5}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{5}}{2*3}=\frac{6+6\sqrt{5}}{6}
$